# Transfrom a Legendre polynomial from $\int_{-1}^{1}\phi_j(x)\phi_k(x)dx$ into $\int_{a}^{b}\phi_j(t)\phi_k(t)dt$ given $t=\dfrac{1}{2}[(b-a)x+(a+b)]$

The Legendre polynomials satisfy

$$\int_{-1}^{1}\phi_j(x)\phi_k(x)dx = \begin{cases} 0 &j\neq k\\\\ \dfrac{2}{2j+1} &j=k \end{cases}$$

Suppose that the best fit problem is given on the interval $[a,b]$. Show that with the transformation $t=\dfrac{1}{2}[(b-a)x+(a+b)]$ and a slight change of notation, we have $$\int_{a}^{b}\phi_j(t)\phi_k(t)dt = \begin{cases} 0 &j\neq k\\\\ \dfrac{b-a}{2j+1} &j=k \end{cases}$$

If $t=\dfrac{1}{2}[(b-a)x+(a+b)]$, then $x=\dfrac{2t-a-b}{b-a}$, and I am stuck at this step, can anyone give me some hints?

• Welcome to scicomp. If this is a homework question, you should tag it as homework... – Jan Oct 22 '13 at 10:32

## 1 Answer

You need to apply the Jacobian of the variable transformation. In other words, since you know how $x$ and $t$ are related, you can figure out how $dx$ and $dt$ are related. Obviously, they should differ by a factor of $(b-a)/2$ since that's the only substantive difference between the two expressions. In terms of more basic calculus, this is simply a $u$-substitution.